Theorem disji 5130

Description: Property of a disjoint collection: if 𝐵(𝑋) = 𝐶 and 𝐵(𝑌) = 𝐷 have a common element 𝑍, then 𝑋 = 𝑌. (Contributed by Mario Carneiro, 14-Nov-2016.)

Hypotheses

Ref	Expression
disji.1	⊢ (𝑥 = 𝑋 → 𝐵 = 𝐶)
disji.2	⊢ (𝑥 = 𝑌 → 𝐵 = 𝐷)

Assertion

Ref	Expression
disji	⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ (𝑍 ∈ 𝐶 ∧ 𝑍 ∈ 𝐷)) → 𝑋 = 𝑌)

Distinct variable groups:   𝑥,𝐴   𝑥,𝐶   𝑥,𝐷   𝑥,𝑋   𝑥,𝑌

Allowed substitution hints:   𝐵(𝑥)   𝑍(𝑥)

Proof of Theorem disji

Step	Hyp	Ref	Expression
1		inelcm 4463	. 2 ⊢ ((𝑍 ∈ 𝐶 ∧ 𝑍 ∈ 𝐷) → (𝐶 ∩ 𝐷) ≠ ∅)
2		disji.1	. . . . . 6 ⊢ (𝑥 = 𝑋 → 𝐵 = 𝐶)
3		disji.2	. . . . . 6 ⊢ (𝑥 = 𝑌 → 𝐵 = 𝐷)
4	2, 3	disji2 5129	. . . . 5 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ 𝑋 ≠ 𝑌) → (𝐶 ∩ 𝐷) = ∅)
5	4	3expia 1119	. . . 4 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → (𝑋 ≠ 𝑌 → (𝐶 ∩ 𝐷) = ∅))
6	5	necon1d 2960	. . 3 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → ((𝐶 ∩ 𝐷) ≠ ∅ → 𝑋 = 𝑌))
7	6	3impia 1115	. 2 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ (𝐶 ∩ 𝐷) ≠ ∅) → 𝑋 = 𝑌)
8	1, 7	syl3an3 1163	1 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ (𝑍 ∈ 𝐶 ∧ 𝑍 ∈ 𝐷)) → 𝑋 = 𝑌)

Syntax hints:   → wi 4   ∧ wa 394   ∧ w3a 1085   = wceq 1539   ∈ wcel 2104   ≠ wne 2938   ∩ cin 3946  ∅c0 4321  Disj wdisj 5112

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1911  ax-6 1969  ax-7 2009  ax-8 2106  ax-9 2114  ax-10 2135  ax-11 2152  ax-12 2169  ax-ext 2701

This theorem depends on definitions:  df-bi 206  df-an 395  df-or 844  df-3an 1087  df-tru 1542  df-fal 1552  df-ex 1780  df-nf 1784  df-sb 2066  df-mo 2532  df-clab 2708  df-cleq 2722  df-clel 2808  df-nfc 2883  df-ne 2939  df-ral 3060  df-rmo 3374  df-rab 3431  df-v 3474  df-sbc 3777  df-csb 3893  df-dif 3950  df-in 3954  df-nul 4322  df-disj 5113

This theorem is referenced by:  volfiniun  25296  fnpreimac  32163